Exercise 5.7.21* (The unbounded component of $\mathscr{C}_f(\mathbb{R})$ is a subgroup of index $2$)

Question

Let $\mathscr{C}_f(\mathbb{R})$ be defined by (5.50) where $a,b,c$ are real numbers, and suppose that the the polynomial on the right side of (5.50) has three real roots $r_1<r_2<r_3$. Let $\mathscr{C}_0$ be the connected component of points $(x,y) \in \mathscr{C}_f(\mathbb{R})$ for which $x \geq r_3$, including the point at infinity, and let $\mathscr{C}_1$ denote the connected component of points for which $r_1\leq x \leq r_2$. Let $P$ and $Q$ be arbitrary points of $\mathscr{C}_f(\mathbb{R})$ . Show that $P+Q$ lies on $\mathscr{C}_0$, or  $\mathscr{C}_1$, according as $P$ and $Q$ lie on the same, or different, components (That is, $\mathscr{C}_0$ is a subgroup of index $2$ in $\mathscr{C}_f(\mathbb{R})$.)

Richard Ganaye · Accepted Answer

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\begin{proof} 
Let $\mathscr{C}_f(\mathbb{R})$ be defined by $y^2 = p(x)$, where 
\begin{align*}
p(x) = x^3 + ax^2 +bx + c = (x-r_1)(x-r_2)(x-r_3),\qquad (r_1,r_2,r_3) \in \R^3,\ r_1<r_2<r_3.
\end{align*}
(thus $-a = r_1+r_2+r_3$).

Then $\mathscr{C}_f(\mathbb{R})$ is the union of the graph of $h$ and of the graph of $-h$, where
$$
h
\left\{
\begin{array}{ccl}
 [r_1, r_2] \cup [r_3,+\infty[ & 	o & \R\
 x & \mapsto & \sqrt{p(x)} = \sqrt{ x^3 + ax^2 +bx + c} = \sqrt{(x-r_1)(x-r_2)(x-r_3)}.
 \end{array}
 ight.
$$
Let $\mathscr{C}_0$ be the connected component of points $(x,y) \in \mathscr{C}_f(\mathbb{R})$ for which $x \geq r_3$, including the point at infinity, and let $\mathscr{C}_1$ denote the connected component of points for which $r_1\leq x \leq r_2$. Since $h$ is continuous, $h([a,b])$ is bounded, so $\mathscr{C}_1$ is the bounded connected component, and $\mathscr{C}_0$ is the unbounded component of $\mathscr{C}_f(\mathbb{R})$.

\bigskip
Let $P= (x_1,y_1), Q = (x_2, y_2)$ be arbitrary points of $\mathscr{C}_f(\mathbb{R})$. We assume without loss of generality that $x_1 \leq x_2$. If $P+Q = (x_3,y_3)$, by the explicit formulas (5.52), (5.53), we have
$$x_3 = \lambda^2 - a - x_1 -x_2,$$
where $\lambda = \frac{y_2 - y_1}{x_2 - x_1}$ if $P 
e Q$, and $\lambda = \frac{p'(x_1)}{2y_1}$ if $P = Q$.

\bigskip
Suppose first that $P$ and $Q$ are on $\mathscr{C}_1$, so that $r_1 \leq x_1 \leq x_2 \leq r_2$.
It is not easy to prove directly that $x_3 \geq r_3$, so we prove this fact indirectly.

\bigskip
{\bf Lemma 1.} {\it Every (projective) line cuts the unbounded component $\mathscr{C}_0$.}

\bigskip
{\it Proof of Lemma 1. } Let $L$ be any line of the plane. If $L$ is a vertical line, with equation $x = k$, and homogeneous equation $x = kt$, then the point at infinity $O = (0:1:0)$ is at the intersection of $L$ and $\mathscr{C}_0$. We suppose now that $L$ has equation $y = mx + p$, and we show that there exists $x_0\geq r_3$ such that $f(x_0, mx_0 + p) = 0$.

Consider the function $\varphi: \R 	o \R$ defined by $\varphi(x) = f(x, mx + p)$. Then
$$\varphi(r_3) = f(r_3, mr_3 + p) =(mr_3 + p)^2 \geq 0.$$
Since $\varphi(x) = (mx +p)^2 - (x-r_1)(x-r_2)(x-r_3) \underset{+\infty}{\sim} -x^3,$ 
$$\lim_{x 	o + \infty} \varphi(x) = -\infty,$$
so there is some $r' \in \R$ such that $\varphi(r') <0$.

Since $\varphi$ is continuous on $\R$, and $\varphi(r_3)\geq 0$, $\varphi(r') < 0$, by the Intermediate Value Theorem, there is some $x_0 \geq r_3$ such that $\varphi(x_0) = 0$, so $f(x_0, mx_0 + p) = 0$. Then the point $M_0 = (x_0, mx_0 +p) \in L \cap \mathscr{C}_0$. \endproof

\bigskip
Now consider the line $L$ passing by the points $P = (x_1, y_1)$ and $Q = (x_2,y_2)$ on $\mathscr{C}_1$ ($L$ is the tangent at $P$ if $P = Q$). By the Lemma, there is a point $R = (x_3,y_3) \in L \cap \mathscr{C}_0$. But a line has exactly three intersection points with the elliptic curve $\mathscr{C}_f(\R)$ (counting multiplicity), so $\mathscr{C}_f(\R) \cap L = \{P,Q,R\}$. This shows that $P + Q + R = 0$, thus $P + Q = -R \in \mathscr{C}_0$. This shows that
\begin{align}
(P \in \mathscr{C}_1 	ext{ and } Q \in \mathscr{C}_1) \Rightarrow P+Q \in \mathscr{C}_0.
\end{align}

\bigskip
{\bf Lemma 2.} { \it Let $A = (r_1,0) \in \mathscr{C}_1$. For every point $M \in \mathscr{C}_0$, there is a unique $N \in \mathscr{C}_1$ such that $M=  A + N$.}

\bigskip
{\it Proof of Lemma 2. } Consider the point $A = (X_1, Y_1) = (r_1,0) \in \mathscr{C}_1$. Then $2A = O$. Consider now any point $M = (X_2,Y_2) \in \mathscr{C}_0$, and $N = A + M= (X_3,Y_3)$. We show first that $N \in \mathscr{C}_1$. By the explicit formula (5.52), using $-a = r_1+r_2 + r_3$ and $X_1 = r_1$,
\begin{align*}
X_3 &= \left(\frac{Y_2-Y_1}{X_2 - X_1} ight)^2 - a - X_1 - X_2\
&=\frac{Y_2^2}{(X_2 -r_1)^2 } + r_2 + r_3  -X_2\
&= r_2 + \frac{Y_2^2 - (X_2 - r_1)^2(X_2 -r_3)}{(X_2 - r_1)^2}\
&=r_2 + \frac{(X_2-r_1)(X_2-r_2)(X_2 - r_3) - (X_2 - r_1)^2(X_2 -r_3)}{(X_2 - r_1)^2}\
&=r_2 - \frac{(r_2 - r_1)(X_2 - r_3)}{X_2 - r_1}\
&\leq r_2,
\end{align*}
because $r_1 < r_2 < r_3 \leq  X_2$. Therefore $X_3 \leq r_2$, so $N = A+ M \in \mathscr{C}_1$, and $A + N = 2A + M = M$. This shows that every point $M \in \mathscr{C}_0$ is of the form $A + N, N \in \mathscr{C}_1$. Moreover $N$ is unique, because $M = A + N = A+ N'$ implies $N = A + M = N'$.  \endproof

\bigskip
We suppose now that $P= (x_1,y_1)$ and $Q = (x_2,y_2)$ are on $\mathscr{C}_0$, and we want to show that $P + Q \in \mathscr{C}_0$. By Lemma 2, there are points $P_1, Q_1$ on $\mathscr{C}_1$ such that $P = A +P_1,\ Q = A + Q_1$. Then $P + Q = (A + P_1) + (A + Q_1) = 2A + P_1 + Q_1 = P_1 + Q_1 \in \mathscr{C_0}$ by (1). We have shown
\begin{align}
(P \in \mathscr{C}_0 	ext{ and } Q \in \mathscr{C}_0) \Rightarrow P+Q \in \mathscr{C}_0.
\end{align}

Finally, if $P \in \mathscr{C}_0$ and $Q \in \mathscr{C}_1$, then by Lemma 2, $P = A + P_1$, where $P_1 \in \mathscr{C}_1$, and $P + Q = A + P_1 + Q = A+R$, where $R = P_1 + Q \in \mathscr{C}_0$ by (1), thus $P+Q = A +R \in \mathscr{C_0}$ by (2). So
\begin{align}
(P \in \mathscr{C}_0 	ext{ and } Q \in \mathscr{C}_1) \Rightarrow P+Q \in \mathscr{C}_1.
\end{align}
We have shown that $P+Q$ lies on $\mathscr{C}_0$, or  $\mathscr{C}_1$, according as $P$ and $Q$ lie on the same, or different, components.

Moreover, if $P = (x,y)$ is on $\mathscr{C}_0$, then $-P = (x, -y) \in \mathscr{C}_0$. This shows with (2) that $\mathscr{C}_0 
e \varnothing$ is a subgroup of $\mathscr{C}_f(\R)$.

By Lemma 2 and (3), $\mathscr{C}_1 = A + \mathscr{C}_0$, therefore
$$\mathscr{C}_f(\R) = \mathscr{C}_0 \cup (A + \mathscr{C}_0),$$
so $\mathscr{C}_0$ is a subgroup of index $2$ in $\mathscr{C}_f(\mathbb{R})$.
\end{proof}

Exercise 5.7.21* (The unbounded component of $\mathscr{C}_f(\mathbb{R})$ is a subgroup of index $2$)

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Exercise 5.7.21* (The unbounded component of $\mathscr{C}_f(\mathbb{R})$ is a subgroup of index $2$)

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